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Arrange the following in order of property indicated for each set : (i) F_(2), Cl_(2), Br_(3), I_(2) - increasing bond dissociation enthalpy. (ii) HF, HCl, HBr, HI - increasing acid strength. (iii) NH_(3), PH_(3), AsH_(3), SbH_(3), BiH_(3) - increasing base strength. |
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Answer» Solution :(i) Bond dissociation enthalpy should decrease as the bond distance INCREASES from `F_(2) "to" I_(2)` due to the corresponding increase in the size of the atom as we move from F to I. However, the F-F bond dissociation enthalpy is smaller than that of Cl-Cl and even smaller than that of BR-Br. This is due to the reason that the F atom is very small and hence the three lone pairs of electrons on each F atom repel the bond pair holding the F-atoms in `F_(2)` molecules. Thus, the bond dissociation enthalpy increases in the order : `I_(2) lt F_(2) lt Br_(2) lt Cl_(2)` (ii) The relative acid strength of HF, HCl, HBr and HI DEPENDS upon their bond dissociation enthalpies. Since the bond dissociation enthalpy, of H-X bond decreases from H-F to H-I as the size of atom increases from F to I. Therefore, the acid strength increases in the opposite order, i.e., the acid strength increases in the order : HF lt HCl lt HBr lt HI. (iii) Due to the presence of a lene pair of electrons on the central atom in `NH_(3), PH_(3), AsH_(3), SbH_(3) and BiH_(3)`, all behave as Lewis BASES. However, as we move from `NH_(3)` to `BiH_(3)`, the size of the atom increases. As a result, the lone pair of electrons occupies a larger volume. In other words, the electron density on the central atom decreases and hence the basic strength decreases as we move from `NH_(3)` to `BiH_(3)` Thus the basic strength increases in the order : `BiH_(3) lt SbH_(3) lt AsH_(3) lt PH_(3) lt NH_(3)`. |
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