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The HNH angle value is higher than HPH, HAsH and HSbH angles. Why ? |
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Answer» <P> Solution :In all the hydrides of group 15 elements, the CENTRAL atom is `sp^(3)`-hybridized. There of the four `sp^(3)`-orbitals FORM three E-H, `sigma-"bonds"` while the fourth CONTAINS the lone pair of electrons as shown below :Since the lone pair-bond pair repulsions are stronger than the bond pair-bond pair repulsions, therefore, in `NH_(3)` the bond angle decreases from `109^(@)-28` to `107.8^(@)`. As we move from N to P to As to Sb, the electronegativity of the central atom goes on decreasing. As a result, bond pairs of electrons, LIE electrons, lie away and away from the central atom. In other words, force of repulsion between the adjacent bond pairs goes on decreasing and consequently the bond angles keep on decreasing from `NH_(3)"to"SbH_(3)`. In other words, HNH bond angle is maximum `(107.8^(@))` followed by HPH `(93.6^(@))`, followed by HAsH`(91.3^(@))`. 
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