although geometries of Nh3 and H2O molecules are distorted tetrahedral bond angle in water is less than that of ammonia, why
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In NH3 ,the N atom is surrounded by three bond pairs (N—H) & ONE lone pair of electrons, which is present on the N atom. The repulsion shown by one lone pair in NH3 towards the three bond pair is very less because the number of bonding electrons is more as compared to lone pair of electron, as a result there will be less repulsion between the lone pair & bonding electrons.So, the bond angle is 107°,nearer to normal tetrahedral bond angle of 109°.5’.
But in case of H2O, the O atom is surrounded by two bond pairs & TWO lone pair of electrons.Consequently ,the repulsion will be stronger & greater between the lone pairs & bond pairs.The bond pairs will be repelled more by the lone pairs ,resulting in decrease of bond angle to 104°, as compared to normal tetrahedral bond angle 109°.5’.
Therefore instead NH3 & H2O have tetrahedral geometry, still they have different bond angles
But in case of H2O, the O atom is surrounded by two bond pairs & TWO lone pair of electrons.Consequently ,the repulsion will be stronger & greater between the lone pairs & bond pairs.The bond pairs will be repelled more by the lone pairs ,resulting in decrease of bond angle to 104°, as compared to normal tetrahedral bond angle 109°.5’.
Therefore instead NH3 & H2O have tetrahedral geometry, still they have different bond angles
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