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In CH_(4),NH_(3) and H_(2)O, the central atom undergoes sp^(3) hybridisation - yet their bond angles are different. Why ? |
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Answer» Solution :(i) In `CH_(4),NH_(3)` and `H_(2)O` the central atom undergoes `sp^(3)` hybridisation. But their bond angles are different DUE to the presence of lone pair of electrons. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs are not same. (iii) Bond pair - Bond `lt` Bond pair - Lone pair `lt` Lone pair - Lone pair so due to the varying repulsive force the bond pairs and lone pairs are distored from regular geometry and organise themselves in such a way that repultion will be minimum and stabily will be MAXIMUM. (iv) In case of `CH_(4)`, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle `=190^(@)28.` `H_(2)O` has 2 bond pairs and 2 lone pairs. There is large repsultion between lp-lp. Again repultion between lp-bp is more than that of 2 bond pairs. So 2 bond are more restricted to form inverted V shape (or) bent shape molecule with a bond angle of `104^(@)35.` (vi) `NH_(3)` has 3 bond pairs and 1 lone pair. There is repulsion between lp-bp. So 3 bonds are more retricted to form pyramidal shape with bond angle equal to `107^(@)18.` |
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