Question

a. Would you except the hydrides of `N,O` and `F` to have lower boiling points than the hydrides of their subsequent group members? Give reason.
b. Can phosphorous with outer electronic configuration `3s^(2)3p^(3)` form `PH_(5)`?
c. How many hydrogen-bonded water molecules(s) are associated with `CuSO_(4).5H_(2)O`?

Answer 1

(a) On the basis of molecular masses of `NH_(3)`, `H_(2)O` and `HF`, their boiling points are expected to be lower than those of the subsequent group member hydrides. Howerver, due to higher electronegativity of `N,O` and `F`, the magnitude of hydrogen bonding in their hydrides will be quite appreciable. hence, the boiling points of `NH_(3),H_(2)O` and `HF` will be higher than the hydride of their subsequent group members.
(b) Although phosphorous exhibits `+3` and `+5` oxidation states, it cannot from `PH_(5)`. besides some other considerations, high `Delta_(a)H` value of dihydrogen and `Delta_(eg)H` value of hydrogen do not favour to exhibit the highest oxidation state of `P` and consequently the formation of `PH_(5)`.
(c) Copper sulphate, `CuSO_(4).5H_(2)O` can be represented as `[Cu(H_(2))O_(4)]^(2+)SO_(4)^(2-).H_(2)O` Only one water molecules, which is present outside the coordination sphere, i.e., square bracket is hydrogen bonded. the other four water molecules are bonded to `Cu^(2+)` by coordinate bond.