In CH4 , NH3 and H2O, the central atom undergoes sp3 hybridisation-yet their bond angles differ. Why?

Question

Answer

Because of lonepairs ......

Answer

It is due to the significant repulsion from lone pairs that there bond angle reduces

Answer

It is due to lone pair present on central atom
bond angle is directly proportinal to electronegativity of central atom
central more electronegative bond angle more if surrounding atoms are same in all compound as it is hydrogen in this case

Answer

It is because of the difference in lone pairs CH4 has no longer,nNH3has 1 LP,nH2o has 2 lone pair

Answer

Introduction:
In CH4, NH3, and H2O molecules, the central atom undergoes sp3 hybridization. This hybridization results in the formation of four sp3 hybrid orbitals that are arranged in a tetrahedral geometry around the central atom. However, despite having the same hybridization, their bond angles differ due to various factors.

Factors affecting bond angles:
1. Lone pairs: The presence of lone pairs on the central atom affects the bond angles. Lone pairs exert a greater repulsion compared to bonded electron pairs, causing the bond angles to deviate from the ideal tetrahedral angle of 109.5 degrees.

2. Steric effects: Steric effects occur when bulky groups or atoms are present in the molecule. These bulky groups can cause repulsion and push the bonded atoms closer together, resulting in a decrease in bond angles.

Explanation for each molecule:
1. CH4 (Methane):
In methane, the central atom carbon is bonded to four hydrogen atoms. It undergoes sp3 hybridization, forming four sp3 hybrid orbitals. These orbitals arrange themselves in a tetrahedral geometry around carbon. The bond angles in methane are approximately 109.5 degrees. The bond angles are close to the ideal tetrahedral angle because there are no lone pairs present on the central carbon atom.

2. NH3 (Ammonia):
In ammonia, the central atom nitrogen is bonded to three hydrogen atoms and has one lone pair. Nitrogen undergoes sp3 hybridization, resulting in the formation of four sp3 hybrid orbitals. However, due to the presence of a lone pair, the bond angles deviate from the ideal tetrahedral angle. The lone pair occupies more space around the nitrogen atom, causing greater repulsion with the bonded electron pairs. As a result, the bond angles in ammonia are approximately 107 degrees, which is slightly less than the ideal tetrahedral angle.

3. H2O (Water):
In water, the central atom oxygen is bonded to two hydrogen atoms and has two lone pairs. Oxygen undergoes sp3 hybridization, forming four sp3 hybrid orbitals. The presence of two lone pairs on oxygen leads to more significant repulsion compared to ammonia. The lone pairs push the bonded electron pairs closer together, resulting in a further decrease in bond angles. The bond angles in water are approximately 104.5 degrees, which is significantly less than the ideal tetrahedral angle.

Conclusion:
In summary, although CH4, NH3, and H2O molecules have the same sp3 hybridization for their central atoms, their bond angles differ due to the presence of lone pairs and steric effects. Lone pairs and bulky groups exert repulsion, leading to a deviation from the ideal tetrahedral angle. The bond angles decrease as the repulsion from lone pairs increases, resulting in different bond angles in each molecule.

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