Explain why bond angle of water is 104' and bond angle of NH3 is 107' both less than tetrahedral bond angle? Explain
Look at the Lewis dot structure. H2O has two sets of lone pairs of electrons and NH3 has one set. Lone pairs tend to occupy more space; therefore, the lone pairs push the remaining atoms (H atoms in this case) closer together in H2O and NH3 so the angle is less than the 109 and 28' of the tetrahedral angle.
You (Abhishek) seem to be using ' (arc minutes) where the units should be degrees
The bond angle in a molecule is determined by the spatial arrangement of its atoms and the repulsion between the electron pairs surrounding the central atom. In the case of water (H2O) and ammonia (NH3), both molecules deviate slightly from the ideal tetrahedral bond angle of 109.5°.
The reason why the bond angles in water (104°) and ammonia (107°) are less than the tetrahedral angle can be explained by the concept of electron pair repulsion theory (VSEPR theory).
In water, the central oxygen atom has two lone pairs of electrons and two bonding pairs. The lone pairs occupy more space than the bonding pairs, resulting in increased electron-electron repulsion between them. This repulsion pushes the bonding electron pairs closer together, which leads to a decrease in the bond angle. As a result, the bond angle in water is reduced to 104°.
Similarly, in ammonia, the central nitrogen atom also has one lone pair and three bonding pairs. Again, the lone pair occupies more space than the bonding pairs, causing increased repulsion between them. This repulsion squeezes the bonding pairs together, resulting in a smaller bond angle of 107°.
The reason both water and ammonia have smaller bond angles compared to the ideal tetrahedral angle is due to the stronger repulsion exerted by the lone pairs of electrons. These lone pairs have a greater influence on the molecular geometry, causing the bond angles to deviate from the perfect tetrahedral arrangement.
The bond angles in water (H2O) and ammonia (NH3) are less than the tetrahedral bond angle because of the lone pairs present on the central atom.
In water, the central oxygen atom has two lone pairs and two bonded hydrogen atoms. The electronic structure of oxygen atom is sp3 hybridized, which means it has one s orbital and three p orbitals available for bonding. The bond angle in a perfect tetrahedron is 109.5° when all four electron domains (lone pairs + bonded atoms) are identical. However, the presence of the two lone pairs in water affects the bond angle. As lone pairs exert greater repulsion than bonded pairs, they push the bonded pairs closer together. This compression of bond angles results in a slightly smaller bond angle of approximately 104°. The repulsion between the lone pairs and the bonded pairs creates an asymmetrical electron arrangement, leading to a distorted tetrahedral geometry.
Similarly, in ammonia (NH3), the nitrogen atom has three bonded hydrogen atoms and one lone pair. The nitrogen atom is also sp3 hybridized, and the lone pair on the central nitrogen exerts greater repulsion than the bonded pairs. The increased repulsion causes the hydrogen-nitrogen-hydrogen bond angle to be slightly smaller than the ideal tetrahedral angle, resulting in an approximate bond angle of 107°.
To determine these bond angles experimentally, various methods such as X-ray crystallography or spectroscopic analysis are used. These techniques allow researchers to measure the angular distances between atoms accurately. In the case of water and ammonia, the experimental data consistently show bond angles smaller than the ideal tetrahedral angle due to the influence of the lone pairs.