The bond angles in a distorted tetrahedral molecule are?
A. Less than 109.6 degrees
B. 109.6 degrees
C. 90 degrees
D. More than 109.6 degrees
The correct answer and explanation is:
The correct answer is A. Less than 109.6 degrees.
In a distorted tetrahedral molecule, the bond angles are typically less than 109.6 degrees. A perfect tetrahedral geometry has bond angles of 109.6 degrees, which is the ideal angle between the bonds in a molecule with four substituents arranged symmetrically around a central atom. However, when the molecule experiences distortion, typically due to factors such as lone pairs of electrons or multiple bonds, the bond angles are reduced from the ideal tetrahedral value.
Lone pairs of electrons take up more space than bonding pairs because lone pairs are localized around the central atom and experience less repulsion from other electron pairs compared to bonding pairs. This results in the repulsion between lone pairs being stronger, which pushes the bonding pairs closer together and decreases the bond angles. Similarly, in the case of multiple bonds or larger substituents around the central atom, these can also cause repulsion that distorts the ideal tetrahedral geometry.
For example, in a molecule like NH3 (ammonia), the nitrogen atom has one lone pair, causing the bond angles between the hydrogen atoms to be less than 109.6 degrees. This effect is seen in many molecules with central atoms having lone pairs, such as H2O (water), where the bond angles are about 104.5 degrees due to two lone pairs on the oxygen atom.
In summary, the bond angles in a distorted tetrahedral molecule are generally less than 109.6 degrees because of the increased repulsion from lone pairs or multiple bonds, which reduces the angles from the ideal tetrahedral configuration.