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In CH_(4),NH_(3) and H_(2)O the central atom undergoes sp hybridization - 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 different due to the presence of lone pair of electrons. (ii) 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 pair < Bond pair - Lone pair < Lone pair - Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and STABILITY 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 = `109^(@)28`’. (v) H.Ohas 2 bond pairs and 2 lone pairs. There is large repulsion between Ip-ip. Again repulsion between Ip-bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (or) bent shape MOLECULE with a bond angle of `104^(@)35`.. (vi) NH, has 3 bond pairs and I lone pair. There is repulsion between Ip-bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107° |
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