The H-C-H bond angle in CH4 is 109.5 degree, the H-N-H bond angle in NH3 is 107 degree, and the H-0-H bond angle in H2O is 104 degree. Explain the trend and why, since the central atom is sp3 hybridized, the angles in ammonia and water are not the ideal 109.5 degree of the tetrahedron.
The Correct Answer and Explanation is :
The trend in bond angles—109.5° in CH₄, 107° in NH₃, and 104° in H₂O—can be explained by the repulsion between electron pairs in the central atom’s valence shell and the varying degree of this repulsion due to the presence of lone pairs. Although all three molecules involve an sp³ hybridized central atom (C in CH₄, N in NH₃, and O in H₂O), the bond angles deviate from the ideal 109.5° tetrahedral angle due to the different numbers of lone pairs on each central atom.
Explanation:
- Methane (CH₄): The central carbon atom is sp³ hybridized and has no lone pairs. This results in an ideal tetrahedral arrangement of the bonding pairs, where the bond angles are exactly 109.5°. Since there are no lone pairs, the repulsion between bonding electron pairs is minimized, leading to the ideal bond angle.
- Ammonia (NH₃): In NH₃, the nitrogen atom is also sp³ hybridized but has one lone pair of electrons. Lone pairs take up more space than bonding pairs because they are localized around the nucleus of the atom, causing greater electron-electron repulsion. As a result, the bond angles in ammonia are slightly compressed to 107°, less than the ideal 109.5°.
- Water (H₂O): The oxygen atom in H₂O is also sp³ hybridized but has two lone pairs of electrons. With two lone pairs, the repulsion between the lone pairs and bonding pairs is even stronger than in ammonia, resulting in a further compression of the bond angles. The H-O-H bond angle in water is 104°, which is significantly smaller than the ideal tetrahedral angle.
Conclusion:
The trend in bond angles reflects the increasing repulsion between lone pairs and bonding pairs as the number of lone pairs on the central atom increases. Lone pairs exert more repulsion than bonding pairs, causing the bond angles to decrease from CH₄ to NH₃ to H₂O.
Let me generate an image showing the molecular geometry of these molecules.
Here is an image showing the molecular geometries of CH₄, NH₃, and H₂O, illustrating the bond angles and lone pairs on the central atoms. It visually represents the differences in bond angles due to the presence of lone pairs in NH₃ and H₂O compared to CH₄.
